Timeline for Arbitrage bounds for Black-Scholes
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2015 at 22:03 | comment | added | Mark Joshi | If the option is out of the money, the price will be positive but it can be arbitrarily close to zero so the best lower bound you can get is zero. There is a whole chapter of my book on this stuff. | |
| Jun 24, 2015 at 17:46 | comment | added | Gordon | @StudentT, Mark"s answer above can also be found in John Hull's book; see Section 11.3 in the 9th edition. | |
| Jun 24, 2015 at 15:32 | comment | added | SmallChess | I don't see it mentioned at least directly in your book. | |
| Jun 24, 2015 at 14:24 | comment | added | Gordon | @emcor, the Jensen inequality is actually strict, then the lower bound is strictly greater than 0. My answer does not emphasize that. | |
| Jun 24, 2015 at 13:40 | comment | added | emcor | @Gordon If its positive, it cannot be zero, so one of you is wrong? | |
| Jun 24, 2015 at 13:33 | comment | added | Gordon | @emcor: What Mark Joshi said that $S_t - Ke^{-r(T-t)} < C$ and positive (for $C$) imply that $\max\big(S_t - Ke^{-r(T-t)}, \, 0\big) < C$, that is, $C > \big(S_t - Ke^{-r(T-t)}\big)^+$. | |
| Jun 24, 2015 at 11:34 | comment | added | emcor | The proof by @Gordon shows that the lower arbitrage bound can be $0$? | |
| Jun 24, 2015 at 10:32 | comment | added | Mark Joshi | at maturity the pay-off satisfies $C_T \geq S_T - K = S_T - KB_T.$ | |
| Jun 24, 2015 at 10:31 | comment | added | Mark Joshi | a negative certainly is a lower bound. I also say that it is positive, | |
| Jun 24, 2015 at 7:41 | comment | added | emcor | The answer is not correct. Let $S<K$. Then your lower bound would be negative. | |
| Jun 24, 2015 at 7:06 | comment | added | emcor | In the first case you say $C\leq S_T$ and then $C\geq S_T-KB_T$. I dont see why this follows directly? | |
| Jun 23, 2015 at 22:29 | history | answered | Mark Joshi | CC BY-SA 3.0 |